Solve the equation 2(x−5)=3x−12(x-5)=3x-1. Answer xx = ... Solve the inequality 7−4y>3(y+2)7-4y>3(y+2). Answer ... Given that 1p−12q=13r\dfrac{1}{p}-\dfrac{1}{2q}=\dfrac{1}{3r}, find pp when q=−1q=-1 and r=2r=2. Answer pp = ... rearrange the formula to make qq the subject. Answer qq = ... Solve the equation $\frac{5}{x+2}-\frac{3x}{2x-1}=3. Give your solutions correct to 2 decimal places. Answer x = ... or ...

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Worked solution

1(a)

1. Expand the bracket and simplify the equation.

2(x−5)=3x−12(x-5)=3x-1

2x−10=3x−12x-10=3x-1

2. Collect like terms and solve for $x$.

2x−10=3x−12x-10=3x-1

−10=x−1-10=x-1

−9=x-9=x

x=−9x=-9

Answer

x=−9x=-9

1(b)

1. Expand the bracket and collect like terms.

7−4y>3(y+2)7-4y>3(y+2)

7−4y>3y+67-4y>3y+6

7−6>3y+4y7-6>3y+4y

2. Solve for $y$ and write the inequality answer.

1>7y1>7y

17>y\dfrac{1}{7}>y

Hence, y<17y<\dfrac{1}{7}.

Answer

y<17y<\dfrac{1}{7}

1(c)(i)

1. Substitute $q=-1$ and $r=2$ into the given formula.

Given
1p−12q=13r\frac{1}{p}-\frac{1}{2q}=\frac{1}{3r}
Substitute q=−1q=-1 and r=2r=2:
1p−12(−1)=13(2)\frac{1}{p}-\frac{1}{2(-1)}=\frac{1}{3(2)}
1p+12=16\frac{1}{p}+\frac{1}{2}=\frac{1}{6}

2. Solve the resulting equation to find $p$.

1p=16−12\frac{1}{p}=\frac{1}{6}-\frac{1}{2}
1p=16−36=−26=−13\frac{1}{p}=\frac{1}{6}-\frac{3}{6}=-\frac{2}{6}=-\frac{1}{3}
Hence
p=−3p=-3

Answer

p=32p=\dfrac{3}{2}

1(c)(ii)

1. Start from the given formula and clear the fractions.

1p−12q=13r\dfrac{1}{p}-\dfrac{1}{2q}=\dfrac{1}{3r}

Multiply throughout by 6pqr6pqr:

6qr−3pr=2pq6qr-3pr=2pq

2. Collect all terms involving $q$ on one side.

6qr−3pr=2pq6qr-3pr=2pq

6qr−2pq=3pr6qr-2pq=3pr

3. Factorise $q$ and divide to make $q$ the subject.

6qr−2pq=3pr6qr-2pq=3pr

q(6r−2p)=3prq(6r-2p)=3pr

q=3pr6r−2p=3pr2(3r−p)q=\dfrac{3pr}{6r-2p}=\dfrac{3pr}{2(3r-p)}

Answer

q=3pr2(3r−p)q=\dfrac{3pr}{2(3r-p)}

1(d)

1. State the restrictions and multiply by the common denominator.

x≠−2x\ne -2 and x≠12x\ne \dfrac12.

Multiply both sides by (x+2)(2x−1)(x+2)(2x-1):
5x+2−3x2x−1=3\frac{5}{x+2}-\frac{3x}{2x-1}=3
5(2x−1)−3x(x+2)=3(x+2)(2x−1).5(2x-1)-3x(x+2)=3(x+2)(2x-1).

2. Expand both sides and simplify to a quadratic equation.

10x−5−3x2−6x=3(2x2+3x−2)10x-5-3x^2-6x=3(2x^2+3x-2)
−3x2+4x−5=6x2+9x−6-3x^2+4x-5=6x^2+9x-6
9x2+5x−1=0.9x^2+5x-1=0.

3. Solve the quadratic equation.

For 9x2+5x−1=09x^2+5x-1=0,
x=−5±52−4(9)(−1)2(9)x=\frac{-5\pm\sqrt{5^2-4(9)(-1)}}{2(9)}
x=−5±6118.x=\frac{-5\pm\sqrt{61}}{18}.

4. Calculate and round the two solutions to 2 decimal places.

x=−5+6118=0.156…≈0.16x=\frac{-5+\sqrt{61}}{18}=0.156\ldots \approx 0.16
x=−5−6118=−0.711…≈−0.71x=\frac{-5-\sqrt{61}}{18}=-0.711\ldots \approx -0.71
Both values are allowed since neither is −2-2 or 12\dfrac12.

Answer

x=−0.71x=-0.71 or x=0.16x=0.16

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